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Difference between revisions of "2008 Indonesia MO Problems/Problem 2"
 
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<cmath> \frac {1}{(1 + \sqrt {x})^{2}} + \frac {1}{(1 + \sqrt {y})^{2}} \ge \frac {2}{x + y + 2}.</cmath>
 
<cmath> \frac {1}{(1 + \sqrt {x})^{2}} + \frac {1}{(1 + \sqrt {y})^{2}} \ge \frac {2}{x + y + 2}.</cmath>
  
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==Solution==
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==Solution 1==
  
 
By the [[Cauchy-Schwarz Inequality]], <math>(1+1)(1+x) \ge (1 + \sqrt{x})^2</math> and <math>(1+1)(1+y) \ge  (1 + \sqrt{y})^2</math>, with equality happening in the earlier inequality when <math>x = 1</math> and equality happening in the latter inequality when <math>y = 1</math>.  Because <math>x,y > 0</math>,
 
By the [[Cauchy-Schwarz Inequality]], <math>(1+1)(1+x) \ge (1 + \sqrt{x})^2</math> and <math>(1+1)(1+y) \ge  (1 + \sqrt{y})^2</math>, with equality happening in the earlier inequality when <math>x = 1</math> and equality happening in the latter inequality when <math>y = 1</math>.  Because <math>x,y > 0</math>,
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Therefore, since <math>\frac{1}{(1 + \sqrt{x})^2} + \frac{1}{(1 + \sqrt{y})^2} \ge \frac12 \cdot (\frac{1}{1+x} + \frac{1}{1+y})</math> and <math>\frac12 \cdot (\frac{1}{1+x} + \frac{1}{1+y}) \ge \sqrt{\frac{1}{(1+x)(1+y)}}</math> and <math>\sqrt{\frac{1}{(x+1)(y+1)}} \ge \frac{2}{x+y+2}</math>, we must have <math>\frac{1}{(1 + \sqrt{x})^2} + \frac{1}{(1 + \sqrt{y})^2} \ge \frac{2}{x+y+2}</math>, with equality happening when <math>x = y = 1</math>.
 
Therefore, since <math>\frac{1}{(1 + \sqrt{x})^2} + \frac{1}{(1 + \sqrt{y})^2} \ge \frac12 \cdot (\frac{1}{1+x} + \frac{1}{1+y})</math> and <math>\frac12 \cdot (\frac{1}{1+x} + \frac{1}{1+y}) \ge \sqrt{\frac{1}{(1+x)(1+y)}}</math> and <math>\sqrt{\frac{1}{(x+1)(y+1)}} \ge \frac{2}{x+y+2}</math>, we must have <math>\frac{1}{(1 + \sqrt{x})^2} + \frac{1}{(1 + \sqrt{y})^2} \ge \frac{2}{x+y+2}</math>, with equality happening when <math>x = y = 1</math>.
  
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Solution:
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==Solution 2==
 
Let <math>f(j)=\frac{1}{(1+\sqrt{j})^2}</math>
 
Let <math>f(j)=\frac{1}{(1+\sqrt{j})^2}</math>
 
Since this function is concave up, according to Jensen's inequality, we can get <math>\frac{f(x)+f(y)}{2}\geq f(\frac{x+y}{2})</math> which means <math>f(x)+f(y)\geq 2f(\frac{x+y}{2})</math>.
 
Since this function is concave up, according to Jensen's inequality, we can get <math>\frac{f(x)+f(y)}{2}\geq f(\frac{x+y}{2})</math> which means <math>f(x)+f(y)\geq 2f(\frac{x+y}{2})</math>.

Latest revision as of 23:09, 3 December 2021

Problem

Prove that for every positive reals $x$ and $y$, \[\frac {1}{(1 + \sqrt {x})^{2}} + \frac {1}{(1 + \sqrt {y})^{2}} \ge \frac {2}{x + y + 2}.\]

Solution 1

By the Cauchy-Schwarz Inequality, $(1+1)(1+x) \ge (1 + \sqrt{x})^2$ and $(1+1)(1+y) \ge (1 + \sqrt{y})^2$, with equality happening in the earlier inequality when $x = 1$ and equality happening in the latter inequality when $y = 1$. Because $x,y > 0$, \begin{align*} \frac{1}{(1 + \sqrt{x})^2} &\ge \frac{1}{2+2x} \\ \frac{1}{(1 + \sqrt{y})^2} &\ge \frac{1}{2+2y} \\ \frac{1}{(1 + \sqrt{x})^2} + \frac{1}{(1 + \sqrt{y})^2} &\ge \frac12 \cdot (\frac{1}{1+x} + \frac{1}{1+y}). \end{align*} By the AM-GM Inequality, we know that $\frac12 \cdot (\frac{1}{1+x} + \frac{1}{1+y}) \ge \sqrt{\frac{1}{(1+x)(1+y)}}$. For the equality case, $\frac{1}{1+x} = \frac{1}{1+y}$, so $x = y$. Additionally, by the AM-GM Inequality, $\frac12 \cdot (\frac{x+1}{2} + \frac{y+1}{2}) \ge \sqrt{\frac{(x+1)(y+1)}{4}}$. For the equality case, $\frac{x+1}{2} = \frac{y+1}{2}$, so $x = y$. Because $x,y \ge 0$, \begin{align*} \frac12 \cdot (\frac{x+y+2}{2}) &\ge \frac{\sqrt{(x+1)(y+1)}}{2} \\ \frac{x+y+2}{2} &\ge \sqrt{(x+1)(y+1)} \\ \sqrt{\frac{1}{(x+1)(y+1)}} &\ge \frac{2}{x+y+2}. \end{align*} Therefore, since $\frac{1}{(1 + \sqrt{x})^2} + \frac{1}{(1 + \sqrt{y})^2} \ge \frac12 \cdot (\frac{1}{1+x} + \frac{1}{1+y})$ and $\frac12 \cdot (\frac{1}{1+x} + \frac{1}{1+y}) \ge \sqrt{\frac{1}{(1+x)(1+y)}}$ and $\sqrt{\frac{1}{(x+1)(y+1)}} \ge \frac{2}{x+y+2}$, we must have $\frac{1}{(1 + \sqrt{x})^2} + \frac{1}{(1 + \sqrt{y})^2} \ge \frac{2}{x+y+2}$, with equality happening when $x = y = 1$.

Solution 2

Let $f(j)=\frac{1}{(1+\sqrt{j})^2}$ Since this function is concave up, according to Jensen's inequality, we can get $\frac{f(x)+f(y)}{2}\geq f(\frac{x+y}{2})$ which means $f(x)+f(y)\geq 2f(\frac{x+y}{2})$. In this problem, it turns into $f(x)+f(y)\geq \frac{2}{(1+\sqrt\frac{x+y}{2})^2}$.The conclusion we try to find is that $f(x)+f(y) \geq \frac{2}{x+y+2}$ So we can see that $\frac{2}{x+y+2} \leq \frac{2}{(1+\sqrt\frac{x+y}{2})^2}$. Take reciprocal for both sides we can get $(x+y+2)\geq (1+\sqrt\frac{x+y}{2})^2$. Take RHS, $(1+\sqrt\frac{x+y}{2})^2=1+\frac{x+y}{2}+\sqrt{2}\sqrt{x+y}$. Now we have to prove that $(x+y+2)\geq (1+\frac{x+y}{2}+\sqrt{2}\sqrt{x+y})$. which turns to $(\frac{x+y}{2}+1)\geq \sqrt{2}\sqrt{x+y}$. It is always correct according to $AM-GM$ inequality, it happens when $x=y=1$. $Q.E.D$ ~bluesoul

See Also

2008 Indonesia MO (Problems)
Preceded by
Problem 1 		1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 Followed by
Problem 3
All Indonesia MO Problems and Solutions
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